What is the integral of (15xe^{2x})/((1+2x)^2) ?

The integral of (15xe^{2x})/((1+2x)^2) is 15(-(e^{2x}x)/(2(1+2x))+1/4 e^{2x})+C

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integral of (15xe^{2x})/((1+2x)^2)

\int \frac{15 x e ^{2 x}}{( 1 + 2 x ) ^{2}} d x

15 (- \frac{e ^{2 x} x}{2 ( 1 + 2 x )} + \frac{1}{4} e^{2 x}) + C

Solution steps

\int \frac{15 x e ^{2 x}}{( 1 + 2 x ) ^{2}} d x

Take the constant out:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= 15 \cdot \int \frac{x e ^{2 x}}{( 1 + 2 x ) ^{2}} d x

Apply Integration By Parts

: - \frac{e ^{2 x} x}{2 ( 1 + 2 x )} - \int - \frac{1}{2} e^{2 x} d x

= 15 (- \frac{e ^{2 x} x}{2 ( 1 + 2 x )} - \int - \frac{1}{2} e^{2 x} d x)

\int - \frac{1}{2} e^{2 x} d x = - \frac{1}{4} e^{2 x}

= 15 (- \frac{e ^{2 x} x}{2 ( 1 + 2 x )} - (- \frac{1}{4} e^{2 x}))

Simplify

= 15 (- \frac{e ^{2 x} x}{2 ( 1 + 2 x )} + \frac{1}{4} e^{2 x})

Add a constant to the solution

= 15 (- \frac{e ^{2 x} x}{2 ( 1 + 2 x )} + \frac{1}{4} e^{2 x}) + C

\frac{d y}{d x}, e^{5 y^{2}} = x^{3} + 3

y^{^{'}} = (x - 2) e^{- 2 y}, y (2) = ln (2)

\frac{\partial}{\partial y} (x e^{\frac{x}{y}})

\int_{25}^{50} 20 d x

f (x) = \frac{x ^{2} + 1}{x + 4}

What is the integral of (15xe^{2x})/((1+2x)^2) ?
The integral of (15xe^{2x})/((1+2x)^2) is 15(-(e^{2x}x)/(2(1+2x))+1/4 e^{2x})+C